Functions Question 445
Question: The function $ f(x)=\sin ( \log (x+\sqrt{x^{2}+1}) ) $ is
[Orissa JEE 2002]
Options:
A) Even function
B) Odd function
C) Neither even nor odd
D) Periodic function
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=\sin ,( \log ,(x+\sqrt{1+x^{2}}) ) $
Þ $ f(-x)=\sin ,[\log ,(-x+\sqrt{1+x^{2}})] $
Þ $ f(-x)=\sin ,\log ( (\sqrt{1+x^{2}}-x)\frac{(\sqrt{1+x^{2}}+x)}{(\sqrt{1+x^{2}}+x)} ) $
Þ $ f(-x)=\sin ,\log ,[ \frac{1}{(x+\sqrt{1+x^{2}})} ] $
Þ $ f(-x)=\sin [ \log {{(x+\sqrt{1+x^{2}})}^{-1}} ] $
Þ $ f(-x)=\sin [ -\log (x+\sqrt{1+x^{2}}) ] $
Þ $ f(-x)=-\sin [ \log (x+\sqrt{1+x^{2}}) ] $
Þ $ f(-x)=-f(x) $ \ $ f(x) $ is odd function.
BETA
